We have
\begin{equation*}
\mathscr{F}\{f(x)\} = \int\limits_{-\infty}^{\infty}e^{-isx}f(x) \,dx
\end{equation*}
Now replace \(f(x)\) by \(f'(x)\) we get -
\begin{equation*}
\mathscr{F}\{f'(x)\} = \int\limits_{-\infty}^{\infty}e^{-isx}f'(x) \,dx
\end{equation*}
\begin{equation*}
= \left.e^{-isx}f(x)\right\vert_{-\infty}^{\infty} + is \int\limits_{-\infty}^{\infty}f(x)e^{-isx}\,dx
\end{equation*}
\begin{equation*}
= 0+is\mathscr{F}\{f(x)\}.
\end{equation*}
[\(\because f(x) \rightarrow 0 \) as \(x\rightarrow \pm \infty\)] Also, generalising these for \(n^{th}\) derivatives, we get -
\begin{equation*}
\mathscr{F}\{f^{n}(x)\} = \mathscr{F}\{\frac{d^{n}f(x)}{\,dx^{n}}\}
\end{equation*}
\begin{equation*}
= \int\limits_{-\infty}^{\infty}\frac{d^{n}f(x)}{\,dx^{n}}e^{-isx}\,dx
\end{equation*}
\begin{equation*}
= \left[\frac{d^{n-1}f(x)}{\,dx^{n-1}}e^{-isx}\right]_{-\infty}^{\infty} - \int\limits_{-\infty}^{\infty}\frac{d^{n-1}f(x)}{\,dx^{n-1}}e^{-isx}(-is)\,dx
\end{equation*}
\begin{equation*}
=0+is\int\limits_{-\infty}^{\infty}\frac{d^{n-1}f(x)}{\,dx^{n-1}}e^{-isx}(-is)\,dx = (is)^{1}\mathscr{F}\{\frac{d^{n-1}f(x)}{\,dx^{n-1}}\}
\end{equation*}
Similarly, we can show that
\begin{equation*}
\mathscr{F}\{\frac{d^{n}f(x)}{\,dx^{n}}\} = (is)^{n}\mathscr{F}\{f(x)\} = (is)^{n} F(s)
\end{equation*}
